Question

Window Help Question 13 474 pts The ionization constant of water at a temperature above 25°C is 3.5 10-14. What is the pH of pure water at this temperature? CALCULATIONS MUST BE PROVIDED FOR FULL CREDIT 2H2O)H30 (aq) + OH (aq) 773 7.00 673 553 1349 6/6 pts Question 14 What is the pH of a solution prepared by dissolving 0.296 g of NaOHis) in 9.00 L of water? CALCULATIONS MUST BE PROVIDED FOR FULL CREDIT 7.000 O2131 3.085 10.915 11.869
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Answer 1

Kwa [Hzot] [ot] [H3ot) = [04] = x x² = 3.5x10-14 x = 1.870 8 2 864 x 10' M = (Hgot] pa-log (1304) = -lag (1-87682869 x 10") = 6.73 Molarity = 0.2969 409 /mod 1000 9000 ml Holarity = 0.00082 2 2 2 H 3 [oH) pOH -log (H) = -log(0.00089222) = 3.085 pH = 14-poh - 14-3.085= 10.915